Question

For a matrix $A = \left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  3&0&2 \\
  1&0&3
\end{array}} \right]$ verify $A\left( {{\text{adj }}A} \right) = \left( {{\text{adj }}A} \right)A = \left| A \right|I$.

Answer 1

Hint: First find the cofactor matrix of A. Determine the cofactors of all elements of matrix A. Then transpose the cofactor matrix to get the adjoint of matrix A, i.e. ${\text{adj}}A$. Multiply the adjoint matrix with A in different order and find the determinant value of A to prove the above result.

Complete step-by-step solution:
According to the question, the given matrix is $A = \left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  3&0&2 \\
  1&0&3
\end{array}} \right]$.
To determine the adjoint of the matrix, first we have to find out the cofactors of all of its elements. Let the cofactors are denoted by $a$ then we have:
Cofactors of first row elements:
\[{a_{1,1}} = \left| {\begin{array}{*{20}{c}}
  0&2 \\
  0&3
\end{array}} \right| = 0\], \[{a_{1,2}} = - \left| {\begin{array}{*{20}{c}}
  3&2 \\
  1&3
\end{array}} \right| = - 7\] and \[{a_{1,3}} = \left| {\begin{array}{*{20}{c}}
  3&0 \\
  1&0
\end{array}} \right| = 0\]
Co-factors of second row elements:
\[{a_{2,1}} = - \left| {\begin{array}{*{20}{c}}
  1&2 \\
  0&3
\end{array}} \right| = - 3\], \[{a_{2,2}} = \left| {\begin{array}{*{20}{c}}
  1&2 \\
  1&3
\end{array}} \right| = 1\] and \[{a_{2,3}} = - \left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&0
\end{array}} \right| = 1\]
Co-factors of third row elements:
\[{a_{3,1}} = \left| {\begin{array}{*{20}{c}}
  1&2 \\
  0&2
\end{array}} \right| = 2\], \[{a_{3,2}} = - \left| {\begin{array}{*{20}{c}}
  1&2 \\
  3&2
\end{array}} \right| = 4\] and \[{a_{3,3}} = \left| {\begin{array}{*{20}{c}}
  1&1 \\
  3&0
\end{array}} \right| = - 3\]
Thus the cofactor matrix is obtained by putting the co-factors in a matrix in order. So we have the cofactor matrix as:
$ \Rightarrow {C_A} = \left[ {\begin{array}{*{20}{c}}
  0&{ - 7}&0 \\
  { - 3}&1&1 \\
  2&4&{ - 3}
\end{array}} \right]$
We know from definition that the adjoint of a matrix is the transpose of its cofactor matrix. So we have the adjoint of matrix A as:
$ \Rightarrow {\text{adj}}A = \left[ {\begin{array}{*{20}{c}}
  0&{ - 3}&2 \\
  { - 7}&1&4 \\
  0&1&{ - 3}
\end{array}} \right]$
To get the value of matrix $A\left( {{\text{adj }}A} \right)$, we will multiply the two matrices, we’ll get:
\[
   \Rightarrow A\left( {{\text{adj }}A} \right) = \left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  3&0&2 \\
  1&0&3
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  0&{ - 3}&2 \\
  { - 7}&1&4 \\
  0&1&{ - 3}
\end{array}} \right] \\
   \Rightarrow A\left( {{\text{adj }}A} \right) = \left[ {\begin{array}{*{20}{c}}
  {0 - 7 + 0}&{ - 3 + 1 + 2}&{2 + 4 - 6} \\
  {0 - 0 + 0}&{ - 9 + 0 + 2}&{6 + 0 - 6} \\
  {0 - 0 + 0}&{ - 3 + 0 + 3}&{2 + 0 - 9}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 7}&0&0 \\
  0&{ - 7}&0 \\
  0&0&{ - 7}
\end{array}} \right] \\
   \Rightarrow A\left( {{\text{adj }}A} \right) = - 7\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]{\text{ }}.....{\text{(1)}} \\
 \]
Similarly to determine the matrix $\left( {{\text{adj }}A} \right)A$, we’ll multiply the matrices in opposite order. We’ll get:
$
   \Rightarrow \left( {{\text{adj }}A} \right)A = \left[ {\begin{array}{*{20}{c}}
  0&{ - 3}&2 \\
  { - 7}&1&4 \\
  0&1&{ - 3}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  3&0&2 \\
  1&0&3
\end{array}} \right] \\
   \Rightarrow \left( {{\text{adj }}A} \right)A = \left[ {\begin{array}{*{20}{c}}
  {0 - 9 + 2}&{0 - 0 + 0}&{0 - 6 + 6} \\
  { - 7 + 3 + 4}&{ - 7 + 0 + 0}&{ - 14 + 2 + 12} \\
  {0 + 3 - 3}&{0 + 0 - 0}&{0 + 2 - 9}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 7}&0&0 \\
  0&{ - 7}&0 \\
  0&0&{ - 7}
\end{array}} \right] \\
   \Rightarrow \left( {{\text{adj }}A} \right)A = - 7\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]{\text{ }}.....{\text{(2)}} \\
 $
Further, the determinant value of matrix A is given as:
$
   \Rightarrow \left| A \right| = 1\left( {0 - 0} \right) - 1\left( {9 - 2} \right) + 2\left( {0 - 0} \right) \\
   \Rightarrow \left| A \right| = 0 - 7 + 0 = - 7 \\
 $
So the value of matrix $\left| A \right|I$ is:
$ \Rightarrow \left| A \right|I - 7\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]{\text{ }}.....{\text{(3)}}$
Here $I = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$ is the identity matrix of order $3 \times 3$.
Now, comparing equations (1), (2) and (3), we can observe that:
$ \Rightarrow A\left( {{\text{adj }}A} \right) = \left( {{\text{adj }}A} \right)A = \left| A \right|I$

Note: This adjoint method is also used to determine the inverse of any matrix. The formula to determine the inverse of any matrix A is given as:
$ \Rightarrow {A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {{\text{adj}}A} \right)$
Thus when all the elements of the adjoint matrix are divided by the determinant value of the original matrix, we get the inverse matrix.